Respuesta :
Answer:
[tex]z=\frac{0.684 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.589[/tex]
[tex]p_v =P(z>1.589)=0.056[/tex]
If we compare the p value obtained with the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65
Step-by-step explanation:
Data given and notation
n=497 represent the random sample taken
X=340 represent the homes in Omaha with one or more lawn mowers
[tex]\hat p=\frac{340}{497}=0.684[/tex] estimated proportion of homes in Omaha with one or more lawn mowers
[tex]p_o=0.65[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the true proportion of homes in Omaha with one or more lawn mowers is higher than 0.65.:
Null hypothesis:[tex]p\leq 0.65[/tex]
Alternative hypothesis:[tex]p > 0.65[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.684 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.589[/tex]
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
[tex]p_v =P(z>1.589)=0.056[/tex]
If we compare the p value obtained with the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65
